416-9sol

# 416-9sol - Denker Spring 2010 416 Stochastic Modeling -...

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Unformatted text preview: Denker Spring 2010 416 Stochastic Modeling - Assignment 9 SOLUTION Problem 1: (Problem 12, p. 347) If X i , i = 1 , 2 , 3, are independent exponential random variables with rates i , respectively, find 1. P ( X 1 &lt; X 2 &lt; X 3 ), 2. P ( X 1 &lt; X 2 | max( X 1 ,X 2 ,X 3 ) = X 3 ), 3. E [max X i | X 1 &lt; X 2 &lt; X 3 ], 4. E [max X i ]. Solution: 1. Let Y = min( X 2 ,X 3 ). Then Y has an exponential distribution with rate = 2 + 3 and is independent of X 1 . But P ( X 1 &lt; Y ) = 1 1 + . Hence P ( X 1 &lt; X 2 &lt; X 3 ) = P ( X 1 = min( X 1 ,X 2 ,X 3 )) P ( X 2 &lt; X 3 | X 1 = min( X 1 ,X 2 ,X 3 )) = 1 1 + 2 + 3 P ( X 2 &lt; X 3 | X 1 = min( X 1 ,X 2 ,X 3 )) . Now we argue as follows: The distribution of X 2- x and X 3- x given that they are larger than X 1 = x is the same as that of X 2 and X 3 , so P ( X 2 &lt; X 3 | min( X 2 ,X 3 ) x ) = P (( X 2- x &lt; X 3- x | min( X 2 ,X 3 ) x ) = P ( X 2 &lt; X 3 ) = 2 2 + 3 . It follows by integrating over x with the distribution of X 1 that P ( X 1 &lt; X 2 &lt; X 3 ) = 1 2 ( 1 + 2 + 3 )( 2 + 3 ) . 2. We can apply the formula from 1. P ( X 1 &lt; X 2 | X 3 = max( X 1 ,X 2 ,X 3 )) = P ( X 1 &lt; X 2 &lt; X 3 ) P ( X 1 &lt; X 2 &lt; X 3 ) + P ( X 2 &lt; X 1 &lt; X 3 ) = 1 1 + 2 + 3 1 + 3 . 3. Let Y = min( X 2 ,X 3 ) and Z = min( X 1 ,X 2 ,X 3 ). Conditioned on { X 1 &lt; X 2 &lt; X 3 } the distribution of X 3- min( X 2 ,X 3 ) is the same as of X 3 : P ( X 3 &gt; t + Y | X 1 &lt; X 2 &lt; X 3 ) = P ( X 3 &gt; t + Y | Y &gt; X 1 ,X 3 &gt; Y ) = integraldisplay P ( X 3 &gt; t + y | y &gt; X 1 ,X 3 &gt; y )( 2 + 3 ) e ( 2 + 3 ) y dy = integraldisplay P ( X 3 &gt; t + y | X 3 &gt; y )( 2 + 3 ) e ( 2 + 3 ) y dy (by independence) = integraldisplay...
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## 416-9sol - Denker Spring 2010 416 Stochastic Modeling -...

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